Hi there!(adsbygoogle = window.adsbygoogle || []).push({});

I 've been thinking about finding the exact value series converge to. How does one do this? I know the convergence proof methods (minor, grater, Leibniz's, quotien, square-root and Cauchy's criteria) - but they just prove the convergence itself and do not give us the exact value.

Here's an example if what I'm talking about

[tex]\displaystyle{\int}e^{-x^2}dx=?[/tex]

[tex]e^x=\displaystyle{\sum_{n=0}^{\infty}}\frac{x^n}{n!}[/tex]

so

[tex]\displaystyle{\int}e^{-x^2} dx=\displaystyle{\int}\displaystyle{\sum_{n=0}^{\infty}}\frac{(-x^2)^n}{n!}dx=\displaystyle{\sum_{n=0}^{\infty}\int}\frac{(-1)^nx^{2n}}{n!}dx[/tex]

[tex]\displaystyle{\int}e^{-x^2} dx=C+\displaystyle{\sum_{n=0}^{\infty}}\frac{(-1)^nx^{2n+1}}{(2n+1)n!}[/tex]

Now this should be the series definition of the Gaussian error function. Suppose C=0 and we're going to take the normalised function for simplicity. How can the following two limits be calculated:

[tex]\displaystyle{\lim_{x\rightarrow +\infty}}\frac{2}{\sqrt{\pi}}\displaystyle{\sum_{n=0}^{\infty}}\frac{(-1)^nx^{2n+1}}{(2n+1)n!}=?(answer: 1)[/tex]

[tex]\displaystyle{\lim_{x\rightarrow -\infty}}\frac{2}{\sqrt{\pi}}\displaystyle{\sum_{n=0}^{\infty}}\frac{(-1)^nx^{2n+1}}{(2n+1)n!}=?(answer: -1)[/tex]

The answers I have seen from Wikipedia: Error function - Wikipedia, the free encyclopedia (cf. the graph)

Thanks in advance, Marine!

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Exact value of convergence

Loading...

Similar Threads for Exact value convergence |
---|

I Derivative of infinitesimal value |

A Integral of Dirac function from 0 to a... value |

I Values of Lagrange multipliers when adding new constraints |

I Finding value of parameters to fit some data |

B Derivation of exact differential |

**Physics Forums | Science Articles, Homework Help, Discussion**